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This is the solution manual of Operations Management by Krajewski.

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128 SYMBOL 108 \f "Wingdings" \s 5 PART 2SYMBOL 108 \f "Wingdings" \s 5Managing Processes

Ch. 1 Competing with Operations. 2, 3, 4 on p27-28

Ch. 2 Project Management. 1, 6, 7, 8,10 on p79-80

Ch. 3 Process Strategy1, 3, 4 on p116

Ch. 4 Process Analysis. 10, 11, 20, 21, 25 on p147-149

Ch. 5. Quality and Performance3, 4, 5, 6, 7, 8, 15 on p206-209

Ch. 6. Capacity Planning2, 3, 4 on p234-235

Ch. 7. Constraint Management3, 4, 5 on p282-283

Ch. 8. Lean Systems3, 4, 7 on p317-318

Ch. 9. Supply Chain Design. 2, 5, 9 on p346-347

Ch. 10 Supply Chain Integration2, 3, 4 on p381-382

Ch. 11. Location. 1, 2, 3, 4, 8, 12 on p407-410

Ch. 12. Inventory Management. 1, 7, 9, 10, 13, 14 on 440-442

Ch. 13. Forecasting.2, 4, 5, 11, 13 on p490-492

Ch. 14. Operations Planning and Scheduling2, 3, 11, 16 on p530-534

Ch. 15. Resource Planning.1, 4, 5, 14 (MRP), 6 (PMS)

Ch12. Suds and Duds Laundry

a. Labor productivity

WeekNumber of

WorkersInput

(Labor-hours)Output

(Shirts)Output/Input

Ratio

1224682.83 shirts/hour

22461302.83 shirts/hour

33621522.45 shirts/hour

43511252.45 shirts/hour

52451312.91 shirts/hour

b. Output per person does not vary much whether it is Sud, Dud, or Jud working. Productivity declines when all three are present. Perhaps there isnt enough work to keep three persons occupied, or perhaps there is not enough work space or equipment to accommodate three workers.3. Compact disc players

Value of Output: $300

Value of Input: Labor + Materials + Overhead

Productivity

10% productivity improvement

Given productivity, and the value of output we solve for the cost of inputs:

Productivity

Input or $136

The cost of inputs must decrease by.

a. A $14 reduction in material costs is

b. A $14 reduction in labor costs is

c. A $14 reduction in overhead is $14/$50 = 28.00%

4. The output of a process is valued at $100 per unit. The cost of labor is $50 per hour including benefits. The accounting department provided the following information about the process for the past four weeks:

Week 1Week 2Week 3Week 4

Units Produced112413101092981

Total Value112,400131,000109,20098,100

Labor ($)12,73514,84210,6039526

Labor (hrs)254.7296.8212.1190.5

Material ($)21,04124,52320,44218,364

Overhead ($)8,99210,4808,7367,848

Multifactor Productivity2.632.632.752.75

Labor Productivity4.41 units/hr4.41units/hr5.15 units/hr5.15 units/hr

a. Use the multifactor productivity ratio to see whether recent process improvements had any effect and, if so, when the effect was noticeable.

Value of output

Value of input: labor + material + overhead

$12,735 + $21,041 + $8,992 = $42,768

Productivity ratio:

Labor Productivity

Week 1Productivity

Week 2Productivity

Week 3Productivity

Week 4Productivity

Improved 4.45% - noticeable in Week 3

b. Has labor productivity changed? Use the labor productivity ratio to support your answer.

Labor-hours of input: Labor $50/hour

Labor costs

Week 1 = $12,735/$50 = 254.7

Week 2 = $14,842/$50 = 296.84

Week 3 = $10,603/$50 = 212.06

Week 4 = $9,526/$50 = 190.52

Productivity ratio:

Labor Productivity

Week 1 = Labor Productivity

Week 2 = Labor Productivity

Week 3 = Labor Productivity

Week 4 = Labor Productivity

Improved 16.68%

Ch2

a. AON network diagram

b. The critical path is ACFHJ with a completion time of 27 days.

EarliestLatestEarliestLatestOn Critical

ActivityDurationStartStartFinishFinishSlackPath?

A2 0 0 2 2 0Yes

B4 2 3 6 7 1No

C5 2 2 7 7 0Yes

D2 615 817 9No

E1 616 71710No

F8 7 71515 0Yes

G3 8171120 9No

H515152020 0Yes

I415161920 1No

J720202727 0Yes

6.

a. The AON diagram is:

b. The critical path is: BEGH, which takes 29 weeks.

c. The slack for activity A = 10 4 = 6 weeks.

The slack for activity D = 13 7 = 6 weeks.

d. If A takes 5 weeks, then D will have 10 5 = 5 weeks slack.

7. Web Ventures Inc.Activity Statistics

ActivityOptimisticMost LikelyPessimisticExpected TimeVariance

(a)(m)(b)()()

A 3 819 97.11

B121518151.00

C 2 616 75.44

D 4 920107.11

E 1 4 7 41.00

8.

a. The expected activity times (in days) are:

ActivityOptimisticMost LikelyPessimistic

A58118.001.00

B48117.831.36

C56 76.000.11

D24 64.000.44

E47107.001.00

PathTotal Expected Time

AC 8 + 6 = 14.00

ADE8 + 4 + 7 = 19.00

BE 7.83 + 7 = 14.83

The critical path is ADE because it has the longest time duration. The expected completion time is 19 days.

b.

Where T = 21 days, = 19 days, and the sum of the variances for critical path ADE is (1.00 + 0.44 + 1.00) = 2.44.

Assuming the normal distribution applies (which is questionable for a sample of three activities), we use the table for the normal probability distribution. Given z=1.28, the probability that the project can be completed in 21 days is 0.8997, or about 90%.

c. Because the normal distribution is symmetrical, the probability the project can be completed in 17 days is (1 0. 8997) = 0. 1003, or about 10%.

10.

a. The AON diagram is:

b. Critical path is BDF. Expected duration of the project is 15 weeks.

c. Activity slacks for the project are:

StartFinishCritical

ActivityEarliestLatestEarliestLatestSlack Path?

A04594No

B00330Yes

C597114No

D33880Yes

E81112153No

F8815150Yes

Ch3

1. Dr. Gulakowicz

Fixed cost, F symbol 61 \f "Symbol" $150,000

Revenue per patient, p symbol 61 \f "Symbol" $3,000

Variable cost per unit, c symbol 61 \f "Symbol" $1000

Break-even volume, patients

3. Baker Machine Company

Closeness

DepartmentRatingCurrent PlanProposed Plan

Pair wijDistance (dij) wijdij Distance (dij)wijdij

1283 243 24

1331 31 3

1591 92 18

1652 101 5

2431 31 3

3582 163 24

3693 272 18

4632 61 3

5631 31 3

wd = 101wd = 101

There is no change in the weighted-distance score. These layouts can be assessed using the Layout solver of OM Explorer, as shown following for the current plan.

4. Baker Machine Block Plan

A good plan would locate the following department pairs close together: 12, 35, 15, 36, 16. The following layout satisfies these requirements and leaves department 3 unmoved. It also provides one-unit distances for department pairs 24 and 46.

36 4

512

The weighted-distance (wd) score is:

[8(l) + 3(2) + 9(l) + 5(l) + 3(l) + 8(l) + 9(l) + 3(l) + 3(2)] = 57, a 43.6% reduction over Problem 3s solution.

Ch4

10. Gasoline Stations

a. The gas station in part (b) has a more efficient flow from the perspective of the customer because traffic moves in only one direction through the system.

b. The gas station in part (a) creates the possibility for a random direction of flow, thereby causing occasional conflicts at the gas pumps.

c. At the gas station in part (b) a customer could pay from the car. However, this practice could be a source of congestion at peak periods.

11. Just Like Home Restaurant

a. The summary of the process chart should appear as follows:

b. Each cycle of making a single-scoop ice cream cone takes

1.70 + 0.80 + 0.25 + 0.50 = 3.25 minutes. The total labor cost is

($10/hr)[(3.25 min/cone)/60 min](10 cones/hr)(10 hr/day)(363 day/yr)

= $19,662.50.

To make this operation more efficient, we can eliminate delay and reduce traveling by having precleaned scoops available. The improved process chart follows.

20. Perrottis Pizza Pareto chart

a. Although the frequency of partly eaten pizza is low, it is a serious quality problem because it is deliberate rather than accidental. It is likely to cause extreme loss of goodwill. A common root cause of many of these problems could be miscommunication between the customer and the order taker, between the order taker and production and between production and distribution. This chart was created using OM Explorer.

b. Cause-and-effect diagram

21. Smith, Schroeder, and Torn (short moves)

a. The tally sheet given in the problem is essentially a horizontal bar chart. To create a Pareto diagram, the categories are arranged in order of decreasing frequency. This diagram was created using OM Explorer.

b. Cause-and-effect diagram

25.Grindwell, Inc.

a. Scatter diagram

b. Correlation coefficient . There is a negative relationship between permeability and carbon content, although it is not too strong.

c. Carbon content must be increased to reduce permeability index.

The cycle time is reduced to 1.65 + 0.45 + 0.25, or 2.35 minutes. The total labor cost is ($ 10/hr)[(2.35 min/cone)/60 min](10 cones/hr)(10 hr/day)(363 day/yr)

= $14,217.50.

Therefore, the annual labor saving is $19,662.50 $14,217.50 = $5,445.00.

Ch5.

3. Garcias Garage

, n = 100, z = 2

At 8 of 100, the number of returns for service is below average, but this observation is within the control limits. The repair process is still in control.

4. Canine Gourmet Company

= 45 grams, n = 10, = 6 grams

a. From Table 5.1,

= 0.308, = 0.223, = 1.777

= 1.777(6 grams) = 10.662 grams

= 0.223(6 grams) = 1.338 grams

= 45 grams + 0.308(6 grams) = 46.848 grams

= 45 grams 0.308(6 grams) = 43.152 grams

b. The range is in statistical control; however, the averages of samples 2, 4, and 5 are out of statistical control, therefore, the process is out of control.

5. Marlin Company

Bottle

Sample1234R

1.604.612.588.600.601.024

2.597.601.607.603.602.010

3.581.570.585.592.582.022

4.620.605.595.588.602.032

5.590.614.608.604.604.024

6.585.583.617.579.591.038

For a quick overview of the data, we can use the Statistics module of POM for Windows, which shows among other things that and ( = 0.0128. The graph tracks the cap diameters over the 6 samples, with four in each sample.

a. , n = 4,

From Table 5.1,

, ,

b. If the process passes the process capability index test, the process is capable.

Process capability index:

Cpk = 1.21

The process is not capable of four-sigma quality. The target is 1.33. Consequently, we test to see if the process variability is too large.

The process variability is below four-sigma quality, which has a target of 1.33. Management and employees should look for ways to reduce the variability in the process and then recheck the process capability index.

6. We initially assume the historical grand average is adequate for the central line of the chart:

Student

Year12345678910Average

16357928770617558637169.7

29077598848836394727074.4

36781935571718698609077.2

46267786189937159938475.7

58588776958909772646076.0

66057798364948664927475.3

79485567789727161929779.4

89786838865877684817181.8

99490768865938687946383.6

108891718997799387698584.9

The average for the process, and the standard deviation of the 100 historical data points in Table 5.2 is 13.

Although the process is in control, the last four observations are all above the average and exhibit an ever-increasing trend. Mega-Byte should explore for causes of corruption, such as instructor or performance measures, which give incentives for improved test scores. It is possible that students are getting brighter or are becoming more highly motivated. Perhaps admissions standards have been raised. It is possible that teaching methods have improved. The point shown here is: the process must be stable while data are collected for setting control limits.

7. Hospital administrator

a. = Total absent/Total observations

= 49/15(64) = 0.051

= 0.0275

= 0.051 + 2.58(0.0275) = 0.1219

= 0.051 2.58(0.0275) = 0.01995, adjusted to zero.

b. The data from the last three weeks fall within the control limits. Therefore we accept the estimate of 5.1% absenteeism. You must now assess whether this amount of absenteeism is typical for nurses aides.

8. Textile manufacturer

a.

b. Because the last two samples with 22 and 21 irregularities plot outside the upper control limit, we conclude that the process is out of control.

15. The Money Pit

a. Lower Specification Calculation

Upper Specification Calculation

b. Because and have values less than 1, the process is not capable of meeting specifications. Yes, valid because the process is under statistical control, as can be shown by plotting the last 15 observations on control charts. Ask students to demonstrate that the process is in statistical control.

c. The variability of the process must be greatly reduced. Also, the process should be better centered between the specification limits.

Ch6.

2. Capacity requirements in five years

This years capacity requirement, allowing instead for just a 5-percent capacity cushion, is 52.63 (or 50 / [1.0 0.05]) customers per day. Essentially you should divide by the desired utilization rate. Five years from now, if demand is only 75 percent of the current level, the customer requirement will be 39.47 (or 52.63 symbol 180 \f "Symbol" \s 12 0.75) customers per day.

3. Airline company

This year's capacity requirement, allowing for a 25-percent capacity cushion, is 93.3 (or 70 / [1.0 symbol 45 \f "Symbol" \s 12 0.25] ) customers per day. Three years from now, if demand increases by 20 percent, the customer requirement will be about 112 (or 93.3 symbol 180 \f "Symbol" \s 12 1.2) customers per day for this flight segment.

4. Automobile brake supplier

a. The total machine hour requirements for all three demand forecasts:

Pessimistic ForecastExpected ForecastOptimistic Forecast

ProcessSetupProcessSetupProcessSetup

ComponentTime

DpTime

(D/Q)sTime

DpTime

(D/Q)sTime

DpTime

(D/Q)s

A750250.0900 300.01,250416.7

B2,000562.52,600 731.33,400956.3

C8501,161.71,2501,708.32,0002,733.3

3,600+ 1,974.24,750+ 2,739.66,650+ 4,106.3

Demand 5,574.27,489.610,756.3

The number of hours (N) provided per machine is:

N = (2 shifts/day symbol 180 \f "Symbol" 8 hours/shift symbol 180 \f "Symbol" 5 days/week symbol 180 \f "Symbol" 52 weeks/year)(1.0 0.2)

= 3,328 hours/machine

The capacity requirements for three forecasts are:

Pessimistic: M = 5,574.2/3328 = 1.67 or 2 machines

Expected: M = 7,489.6/3328 = 2.25 or 3 machines

Optimistic: M = 10,756.3/3328 = 3.23 or 4 machines

b. The current capacity is sufficient for the pessimistic and expected forecasts. However, there is a gap of one machine for the optimistic forecast. The gap drops to zero when the 20 percent increase from short-term options is included.

3 machines symbol 180 \f "Symbol" 3,328 hours/machine symbol 180 \f "Symbol" 1.2 = 11,981 hours.

This is greater than 10,756 hours required.

Ch73. CKC

Station X is the bottleneck 2600 minutes

Work StationProduct AProduct BTotal Load

W10*90=90014*85=11902090

X10*90=90020*85=17002600

Y15*90=135011*85=9352285

4. CKC

a. Traditional Method: Product B has the higher contribution margin/unit

Product AProduct B

Price55.0065.00

Raw and Purchased Parts5.0010.00

Contribution Margin50.0055.00

Work StationMinutes at StartMins. Left after Making 85 BsMins. Left after Making 90 AsCan Only Make 70 As

W24001210310

X2400700700/10 = 70

Y24001465115

85 units of B and 70 units of A (Product B will use 1700 minutes at station X leaving 700 for Product A.

ProductOverheadRaw MatlLaborPurchase PartsTotal CostsRevenues

A70 x 2 =14070 x 3 = 21070 x $55 = 3850

B85 x 5 = 42585 x 5 = 42585 x $65 = 5525

Totals35005653 x $6 x 40 hrs = 72063554209375

Revenue costs = profit

$9,375 - $5,420 = $3,955

b. Bottleneck-based approach: Product A has the higher contribution margin/unit at the bottleneck

Product AProduct B

Margin50.0055.00

Time at bottleneck10 min20 min

Contribution margin per minute5.002.75

Work StationMinutes at StartMins. Left after Making 90 AsMins. Left after Making 85 BsCan Only Make 75 Bs

W24001500310

X240015001500/20 = 75

Y24001050115

Make 90 units of A (900 minutes used leaves 1500 minutes) can make 75 units of B

ProductOverheadRaw MatlLaborPurchase PartsTotal CostsRevenues

A90 x 2 = 18090 x 3 = 27090 x $55 = 4950

B75 x 5 = 37575 x 5 = 37575 x $65 = 4875

Totals3500 5553 x $6 x 40 hrs = 720 6405415 9825

Profit=Revenue costs

$9,825 $5,415 = $4,410

c. $4,410- $3,955 = $455 increase using TOC, which is a 12% increase

5. Student answers will vary - this is one possible solution. Assembly-line balancing with longest work element rule to produce 40 units per hour.

a.

b.

c. S1 = {A, C, E}, S2 = {B}, S3 = {G, D}, S4 = {H, F, I}, S5 = {J, K}

Work ElementCumulativeIdle Time

StationCandidate(s)ChoiceTime (sec)Time (sec)( sec)

S1AA404050

CC307020

EE2090 0

S2BB808010

S3D, F, GG606030

D, F, ID2585 5

S4F, H, IH454545

F, IF156030

II107020

S5JJ757515

KK1590 0

d. Efficiency (%)

Ch8

3. LeWin

a. Solving for implied policy variable, (k =

b. Reduction in waiting time

The reduction in waiting time is:

4. Gadjits and Widjits

a. Containers for gadjits

k =

k = = 4.905

k = 5

b. Containers for widjits

k =

k = = 11.750

k = 12

7. Januarys container needs

k =

k = = 7.16 or 8 containers

Februarys container needs

k =

k = (900*4) (0.16+0.125)(1+0.15)200

k = 5.8995 or 6 containers per day

Ch9.

2. Prince Electronics

a. Value of each DCs pipeline inventory

= (75 units/wk)(2 wk)($350/unit)

= $52,500

b. Total inventory= cycle + safety + pipeline

= 5[(400/2) + (2*75) + (2*75)]

= 2,500 units

5. Precision Enterprises. Average aggregate inventory value

= Raw materials + WIP + Finished goods

= $3,129,500 + $6,237,000 + $2,686,500

= $12,053,000

a. Sales per week= Cost of goods sold/52 weeks per year

= $32,500,000/52

= $625,000

Weeks of supply= Average aggregate inventory value/

Weekly sales

= $12,053,000/$625,000

= 19.28 wk

b. Inventory turnover= (Annual sales at cost)/(Average

aggregate inventory value)

= $32,500,000/$12,053,000

= 2.6964 turns/year

9. Sterling Inc.

a.Average

Part NumberInventory (units)Value ($/unit)Total Value ($)

RM-120,0001.0020,000

RM-25,0005.0025,000

RM-33,0006.0018,000

RM-41,0008.008,000

WIP-16,00010.0060,000

WIP-28,00012.0096,000

FG-11,00065.0065,000

FG-250088.0044,000

Average aggregate inventory value: $336,000

b. Average weekly sales at cost= $6,500,000/52

= $125,000

Weeks of supply = $336,000/$125,000

= 2.688 weeks.

c. Inventory turnover = Annual sales (at cost) /Average aggregate inventory value

= $6,500,000/$336,000

= 19.34 turns.

Ch10

2. Eight Flags. We apply the equation for total annual cost analysis to each supplier:

Total Annual Cost = pD + Freight costs + (Q/2 + L)H + Administrative costs.

The average requirements per week are 30,000/50 = 600 gallons.

For Sharps and a shipping quantity of 5,000, the total annual cost is:

Total Annual Cost = ($4)(30,000) + $5,000 + (5,000/2 + 600 (4))($0.80) + $4,000 = $132,920.

The total annual costs for the other alternatives are given in the following table.

Shipping Quantity

Supplier5,00010,00015,000

Sharps$132,920$132,520$133,920

Winkler$129,136$128,736$130,336

Winkler, with a shipping quantity of 10,000, is the lowest cost alternative.

3. Bennet

a. Each suppliers performance can be calculated as:

PerformanceWeighted Rating

CriterionWeightSupplier ASupplier BSupplier C

1.Price0.20.6(0.2) = 0.120.5(0.2) = 0.100.9(0.2) = 0.18

2.Quality0.20.6(0.2) = 0.120.4(0.2) = 0.080.8(0.2) = 0.16

3.Delivery0.30.6(0.3) = 0.180.3(0.3) = 0.090.8(0.3) = 0.24

4.Production facilities & capacity0.10.5(0.1) = 0.050.9(0.1) = 0.090.6(0.1) = 0.06

5.Environmental protection0.10.7(0.1) = 0.070.8(0.1) = 0.080.6(0.1) = 0.06

6.Financial position0.10.9(0.1) = 0.090.9(0.1) = 0.090.7(0.1) = 0.07

Total weighted score0.630.530.77

b. Suppliers A and C survived the hurdle. Supplier A would receive 45% of the orders and Supplier C would receive 55% of the orders.

c. Bens system provides some assurance that orders are placed with qualified suppliers. The orders are divided between two suppliers, so there is a ready alternative if a strike, fire, or other problem prevents one supplier from performing. The system also rewards suppliers with more orders if they improve performance.

4. Beagle Clothiers. The weights for the four criteriaprice, quality, delivery, and flexibilityshould be 0.2, 0.2, 0.2, and 0.4, respectively. The weighted scores are

Supplier ASupplier BSupplier C

8 symbol 180 \f "Symbol" 0.2 = 1.66 symbol 180 \f "Symbol" 0.2 = 1.26 symbol 180 \f "Symbol" 0.2 = 1.2

9 symbol 180 \f "Symbol" 0.2 = 1.87 symbol 180 \f "Symbol" 0.2 = 1.47 symbol 180 \f "Symbol" 0.2 = 1.4

7 symbol 180 \f "Symbol" 0.2 = 1.49 symbol 180 \f "Symbol" 0.2 = 1.86 symbol 180 \f "Symbol" 0.2 = 1.2

5 symbol 180 \f "Symbol" 0.4 = 2.08 symbol 180 \f "Symbol" 0.4 = 3.29 symbol 180 \f "Symbol" 0.4 = 3.6

Total weighted score6.87.67.4

Supplier B should be selected.

Ch11

1. Preference matrix location for A, B, C, or D

FactorFactor Score for Each Location

Location FactorWeightABCD

1. Labor climate 55 254 203 155 25

2. Quality of life 302 603 9051501 30

3. Transportation system 53 154 203 155 25

4. Proximity to markets 2551253 7541004100

5. Proximity to materials 53 152 103 155 25

6. Taxes 152 305 755 754 60

7. Utilities 155 754 602 301 15

Total100345350400280

Location C, with 400 points.

2. John and Jane Darling

FactorFactor Score for Each Location

Location FactorWeightABCD

1. Rent253751252505125

2. Quality of life2024051005100480

3. Schools531552531515

4. Proximity to work10550330440330

5. Proximity to recreation15460460575230

6. Neighborhood security15230460460460

7. Utilities10440220330550

Total100310320370380

Location D, the in-laws downstairs apartment, is indicated by the highest score. This points out a criticism of the technique: the Darlings did not include or give weight to a relevant factor.

3. Jackson or Dayton locations

Jackson

Dayton

Jackson yields higher total profit per year.

4. Fall-Line, Inc.

a. Plot of total costs (in $ millions) versus volume (in thousands)

b. Medicine Lodge is the lowest-cost location for volumes up to 25,000 pairs per year.

Broken Bow is the best choice over the range of 25,000 to 44,000 pairs per year.

Wounded Knee is the lowest-cost location for volumes over 44,000 pairs per year.

Aspen is not the low-cost location at any volume.

c. Aspen

Medicine Lodge

Broken Bow

Wounded Knee

d. Aspen would surpass Broken Bow when the Aspen profit is $7,780,000.

Aspen would be the best location if sales would exceed 63,120 pairs per year. Holding all other sales volumes constant.

8. Centura High School

Using the Center of Gravity Solver of OM Explorer, we get:

Solver - Center of Gravity

Enter the names of the towns and the coordinates (x and y) and population (or load, l) of each town.

City/Town Namex y l lx ly

Boelus106.7246.3122824332.1610558.68

Cairo106.6846.3773778623.1634174.69

Dannebrog106.7746.3435638010.1216497.04

00

00

1321140965.461230.41

Center-of-Gravity Coordinatesx*106.71

y*46.35

12. Davis, California, Post Office

a. Center of Gravity

and

b.Load distance scores

Mail Source

PointRound Trips

per Day (l)xy-

CoordLoad-distance to

M: (10, 3)Load-distance to

CG: (8.9, 6.5)

16(2, 8)6(8 + 5) = 786(6.9 + 1.5) = 50.4

23(6, 1)3(4 + 2) = 183(2.9 + 5.5) = 25.2

33(8, 5)3(2 + 2) = 123(0.9 + 1.5) = 7.2

43(13, 3)3(3 + 0) = 93(4.1 + 3.5) = 22.8

52(15, 10)2(5 + 7) = 242(6.1 + 3.5) = 19.2

67(6, 14)7(4 + 11) = 1057(2.9 + 7.5) = 72.8

75(18, 1)5(8 + 2) = 505(9.1 + 5.5) = 73.0

M3(10, 3)3(0 + 0) = 03(1.1 + 3.5) = 13.8

Total = 296Total = 284.4

Ch121. Lockwood Industries

First we rank the SKUs from top to bottom on the basis of their dollar usage. Then we partition them into classes. The analysis was done using OM Explorer Tutor12.1ABC Analysis.

Cumulative % Cumulative %

SKU #DescriptionQty Used/YearValueDollar UsagePct of Totalof Dollar Valueof SKUsClass

444,000$1.00$44,00060.0%60.0%12.5%A

770,000$0.30$21,00028.6%88.7%25.0%A

5900$4.50$4,0505.5%94.2%37.5%B

2120,000$0.03$3,6004.9%99.1%50.0%B

6350$0.90$3150.4%99.5%62.5%C

8200$1.50$3000.4%99.9%75.0%C

3100$0.45$450.1%100.0%87.5%C

11,200$0.01$120.0%100.0%100.0%C

Total$73,322

SHAPE \* MERGEFORMAT

The dollar usage percentages dont exactly match the predictions of ABC analysis. For example, Class A SKUs account for 88.7% of the total, rather than 80%. Nonetheless, the important finding is that ABC analysis did find the significant few. For the items sampled, particularly close control is needed for SKUs 4 and 7.

7. Sams Cat Hotel

a. Economic order quantity

= 90/week

D = (90 bags/week)(52 weeks/yr) = 4,680

S = $54

Price = $11.70

H = (27%)($11.70) = $3.16

= 399.93, or 400 bags.

Time between orders, in weeks

b. Reorder point, R

R = demand during protection interval + safety stock

Demand during protection interval = L = 90 * 3 = 270 bags

Safety stock = zdLT

When the desired cycle-service level is 80%, .

= 15 = 25.98 or 26

Safety stock = 0.84 * 26 = 21.82, or 22 bags

c. Initial inventory position = OH + SR BO = 320 + 0 0

320 10 = 310.

Because inventory position remains above 292, it is not yet time to place an order.

d. Annual holding cost Annual ordering cost

When the EOQ is used these two costs are equal. When , the annual holding cost is larger than the ordering cost, therefore Q is too large. Total costs are $789.75 + $505.44 = $1,295.19.

9. A Q system (also known as a reorder point system)

= 300 pints/week

= 15 pints

a. Standard deviation of demand during the protection interval:

= 15 = 45 pints

b. Average demand during the protection interval:

Demand during protection interval = L = 300 * 9 = 2700 pints

c. Reorder point

R = average demand during protection interval + safety stock

Safety stock = zdLT

When the desired cycle-service level is 99%, z = 2.33.

Safety stock = 2.33 * 45 = 104.85 or 105 pints

R = 2,700 + 105 0 = 2,805 pints

a. Annual holding cost Annual ordering cost

Total cost using EOQ is $1,263.60, which is $31.59 less than when the order quantity is 500 bags.

10. Petromax Enterprises

a.

b. Safety stock = zdLT = = (1.28)(125) = 277.13 or 277 units

Reorder point= average lead time demand + safety stock

= (3)(50,000/50) + 277

= 3,277 units

13. Nationwide Auto Parts

a. Protection interval (PI)= P + L = 6 +3 = 9 weeks

Average demand during PI= 9 (100) = 900 units

Standard deviation during PI= = 60 units

b. Target inventory= (P+L) + z(P+L = 900 + (1.96)(60) = 1,018

c. Order quantity= Target inventory IP

= 1,018 350 = 668 units presuming no SR or BO

14. A P system (also known as a periodic review system).

Find cycle-service level, given:

L = 2 weeks

P = 1 week

(P + L) = 218 boxes

= 40 boxes

T = 300 boxes

T = Average demand during protection interval + Safety stock

T = 218 + z(40) = 300 boxesz = (300 218)/40 = 2.05When z = 2.05, cycle-service level is 97.98 or 98%.Ch132. Dalworth Company

a. Three-month simple moving average

MonthActual SalesThree-Month Simple Absolute AbsoluteSquared

(Thousands)Moving AverageError% ErrorError

Forecast

Jan.20

Feb.24

Mar.27

Apr.31

May37(24+27+31)/3 = 27.339.6726.1493.51

June47(27+31+37)/3 = 31.6715.3332.62235.01

July53(31+37+47)/3 = 38.3314.6727.68215.21

Aug.62(37+47+53)/3 = 45.6716.3326.34266.67

Sept.54(47+53+62)/3 = 54.000.000.000.00

Oct.36(53+62+54)/3 = 56.3320.3356.47413.31

Nov.32(62+54+36)/3 = 50.6718.6758.34348.57

Dec.29(54+36+32)/3 = 40.6711.6740.24136.19

Total106.67267.831,708.47

Average13.3333.48213.56

Such results also can be obtained from the Time Series Forecasting Solver of OM Explorer:

b. Four-month simple moving average

MonthActual SalesFour-Month SimpleAbsoluteAbsoluteSquared

(Thousands)Moving AverageError% ErrorError

Forecast

Jan.20

Feb.24

Mar.27

Apr.31

May37(20+24+27+31)/4 = 25.511.5031.08132.25

June47(24+27+31+37)/4 = 29.7517.2536.70297.56

July53(27+31+37+47)/4 = 35.517.5033.02306.25

Aug.62(31+37+47+53)/4 = 42.0020.0032.26400.00

Sept.54(37+47+53+62)/4 = 49.754.257.8718.06

Oct.36(47+53+62+54)/4 = 54.0018.0050.00324.00

Nov.32(53+62+54+36)/4 = 51.2519.2560.16370.56

Dec.29(62+54+36+32)/4 = 46.0017.0058.62289.00

Total124.75309.712,137.68

Average15.5938.71267.21

Similarly, using Time Series Forecasting Solver of OM Explorer, we get:

c.symbol 45 \f "Symbol" \s 12e.Comparison of performance

Question Measure 3-Month 4-Month Recommendation

SMASMA

c.MAD13.3315.593-month SMA

d.MAPE33.4838.713-month SMA

e.MSE213.56267.213-month SMA

4. Dalworth Company (continued)

c. Three-month weighted moving average (weights of 3/6, 2/6, and 1/6)

Month Actual SalesThree-Month WeightedAbsolute Absolute % Squared

(000s)Moving Average ForecastErrorErrorError

Jan.20

Feb.24

Mar.27

Apr.31[(327)+(224)+(l 20)]/6 = 24.836.1719.9038.07

May37[(331)+(227)+(l 24)]/6 = 28.508.5022.9772.25

June47[(337)+(231)+(l 27)]/6 = 33.3313.6729.09186.87

July53[(347)+237)+(l 31)]/6 = 41.0012.0022.64144.00

Aug.62[(353)+(247)+(l 37)]/6 = 48.3313.6722.05186.87

Sept.54[(362)+(253)+(l 47)]/6 = 56.502.504.636.25

Oct.36[(354)+(262)+(l 53)]/6 = 56.5020.5056.94420.25

Nov.32[(336)+(254)+(l62)]/6 = 46.3314.3344.78205.35

Dec.29[(332)+(236)+(l 54)]/6 = 37.008.0027.5964.00

Total99.34250.591,323.91

Average11.0427.84147.09

The results from Time Series Forecasting Solver of OM Explorer give the same results:

d. Exponential smoothing (symbol 97 \f "Symbol" \s 12 = 0.6)

Month DtFtFt+1 = Ft + symbol 97 \f "Symbol" \s 10(Dt symbol 45 \f "Symbol" \s 10 Ft)Absolute Absolute Squared

(t) (millions)(Forecast for Next Month)Error% ErrorError

Jan.2022.00 20.80

Feb.2420.8022.72

Mar.2722.7225.29

Apr.3125.2928.725.7118.4132.60

May3728.7233.698.2822.3868.56

June4733.6941.6713.3128.32177.16

July5341.6748.4711.3321.38128.37

Aug.6248.4756.5913.5321.82183.06

Sept.5456.5955.042.594.806.71

Oct.3655.0443.6219.0452.88362.52

Nov.3243.6236.6411.6136.28134.79

Dec.2936.6432.067.6526.3858.52

Total93.05232.651,152.29

Average 10.3425.85128.03

c.symbol 45 \f "Symbol" \s 12e.Comparison of performance

Question Measure 3-Month Exponential Recommendation

WMA Smoothing

c.MAD 11.0410.34Exponential smoothing

d.MAPE 27.8425.85Exponential smoothing

e.MSE147.09128.03Exponential smoothing

5. Convenience Store

May

June

July

11. Snyders Garden Center

SeasonalSeasonalAverage

QuarterYear 1FactorYear 2FactorSeasonal Factor

1400.179600.2180.199

23501.5734401.6001.587

32901.3033201.1641.234

42100.9442801.0180.981

Total8901,100

Average222.50275.00

Average quarterly sales in year 3 are expected to be 287.50 (1,150/4). Using the average seasonal factors, the forecasts for year 3 are:

QuarterForecast

10.199(287.50) 57

21.587(287.50)456

31.234(287.50)355

40.981(287.50)282

With the Seasonal Forecasting Solver of OM Explorer, we get the same results

13. Garcias Garage

a. The results, using the Regression Analysis Solver of OM Explorer, are:

The regression equation is Y = 42.464 + 2.452X

b. Forecasts

Y (Sep) = 42.464 + 2.452 (9) = 64.532 or 65

Y (Oct) = 42.464 + 2.452 (10) = 66.984 or 67

Y (Nov) = 42.464 + 2.452 (11) = 71.888 or 72

Ch142. Bob Carltons Golf Camp

a. The level strategy:

The peak demand is 6,400 hours in quarter 2. As each employee can work 600hours per quarter (480 on regular time and 120 on overtime), the level workforce that covers requirements and minimizes undertime is 6,400/600=10.67 or 11 employees.CostCalculationAmount

Regular wages($7200 per quarter)(11)(8 quarters)$633,600

Overtime wages*(1,120 hr in quarter 2)($20 per hr)22,400

(960 hr in quarter 6)($20 per hr)19,200

Hire costs($10,000 per hire)(3 hires)30,000

TOTAL$705,200

*The 11 workers can produce (11) (480)=5,280 hours of regular time in any quarter. The 6,400-hour requirement in quarter 2 exceeds this amount by 1,120 hours. The 6,240-hour requirement in quarter 6 exceeds this amount by 960 hours.

The total undertime hours can be calculated as:

Quarter 111(480) 4,2001,080 hours

Quarter 311(480) 3,0002,280

Quarter 411(480) 4,800480

Quarter 511(480) 4,400880

Quarter 711(480) 3,6001,680

Quarter 811(480) 4,800480

6,880 hours

b. The chase strategy:

QuarterDemand (hr)WorkforceHiresLayoffs

14,200 91

26,400145

33,000 77

44,800103

54,40010

66,240133

73,600 85

84,8001020

TOTAL811412

CostCalculationAmount

Regular wages($7,200 per quarter)(81)$583,200

Hire costs($10,000 per hire)(14 hires)140,000

Layoff costs($4,000 per layoff)(12 layoffs)48,000

TOTAL$771,200

c. Proposed plan:

This plan begins with just 9 workers for Quarter 1, as with the chase strategy. However, it increases temporarily the workforce to 11 employees in Quarters 2 and 6, making up the shortfall with overtime.

QuarterDemand (hr)WorkforceHiresLayoffsOvertime (hr)

14,200 91

26,4001121,120

33,000 92

44,800 9 480

54,400 9 80

66,240112 960

73,600 92

84,800 900 480

TOTAL76543,120

CostCalculationAmount

Regular wages($7,200 per quarter)(76)$547,200

Hire costs($10,000 per hire)(5 hires)50,000

Layoff costs($4,000 per layoff)(4 layoffs)16,000

Overtime($20 per hour)(3,120 hours)62,400

TOTAL$675,600

This plan is more like the level strategy, except that only 9 employees are on the workforce each quarter, with another 2 hired temporarily in Quarters 2 and 6. It also uses more overtime than with the level strategy.

3. Bob Carltons Golf Camp with part-time instructors

a. One of many plans that take advantage of flexibility provided by part-time instructors, this plan reduces hiring and layoffs of certified instructors, reduces overtime, and reduces total costs.

DemandCertifiedCertCertPTPTPTOvertime

Qtr(hr)WorkforceHiresLayoffsWork HoursHiresLayoffs(hr)

14,200 91

26,4001017203880

33,000 823

44,800 87203240

54,400 8560

66,240102720720

73,600 823

84,800 87203240

TOTAL69443,440962,080

CostCalculationAmount

Regular wages($7,200 per quarter)(69)$496,800

Cert. hire costs($10,000 per hire)(4 hires)40,000

Cert. layoff costs($4,000 per hire)(4 layoffs)16,000

PT. hire costs($2,000 per hire)(9 hires)18,000

PT. labor costs($12/hr) (3,440 hrs)41,280

Overtime($20 per hour)(2,080 hours)41,600

TOTAL$653,680

11. Michaels Distribution Center

DayMTWThFSSu

Requirements6353723

MTWThFSSuEmployee

63537231

52426232

41315233

31304124

20203125

10202016

00101017

The number of employees is 7. They are scheduled to take the boxed days off.

16. Hickory Company

a. Schedules for two rules

FCFS rule:

Customer SequenceHr Since Order ArrivedStart Time (hr)Machine Time (hr)Finish Time (hr)Due Date(hr)Past Due (hr)Flow Time (hr)

160+10=1012016

2510+3=138518

3313+15=28181031

4128+9=37201738

5037+7=44212344

Average flow time = = 29.4 hours

Average hours past due = = 11.0 hours

EDD rule:

Customer SequenceHr Since Order ArrivedStart Time (hr)Machine Time (hr)Finish Time (hr)Due Date(hr)Hr Past DateFlow Time (hr)

250+3=3808

163+10=1312119

3313+15=28181031

4128+9=37201738

5037+7=44212344

Average flow time = = 28.0 hours

Average hours past due = = 10.2 hours

b. The EDD rule is better than FCFS on both average flow time (28.0 vs. 29.4) and average hours past due (10.2 vs. 11.0). It gives the better schedule, although this is not always true.

Ch15.

1.Bill of materials, Fig. 15.24

a.Item I has only one parent (E). However, item E has two parents (B and C).

b.Item A has 10 unique components (B, C, D, E, F, G, H, I, J, and K).

c.Item A has five purchased items (I, F, G, H, and K). These are the items without components.

d.Item A has five intermediate items (B, C, D, E, and J). These items have both parents and components.

e.The longest path is IECA at 11 weeks.

4.The bill of materials for item A with lead times is shown following.

a.Lead time is determined by the longest path

G-E-B-A = 12 weeks.

b.If purchased items D, F, G, and H are already in inventory, the lead time is reduced to: ABE = 8 weeks.

c.Item G is the purchased item with the longest lead time in the longest path. This purchased item could be kept in stock to reduce the overall lead time.

5.Refer to Figure 15.19 and Solved Problem 1.

FIGURE 15.19

Material requirement to produce 5 units of end-item A:

on hand

Material requirement to produce 13B:

13D

26E

Material requirement to produce 5C:

5D on hand

Material requirement to produce 4F:

4G 3G on hand = 1G

Purchased material requirements net of on-hand inventory: , 26E, and 1G.

6. MPS record in Fig. 15.26. The following table is from the Master Production Scheduling Solver in OM Explorer. The ATP row is not required for this problem.

Solver Master Production Scheduling

Enter data in yellow shaded areas.

Lot Size60

Lead Time1

Quantity on Hand35123456789101112131415

Forecast2018282823303338

Customer Orders (Booked)151791497

Projected On-Hand Inventory15572913883557

MPS Quantity60606060

MPS start60606060

Available-to-Promise Inv (ATP)2020515360

14.Inventory record.

The tables following were generated with the Single-Item MRP solver from OM Explorer.

a.Fixed order quantity = 110

b.L4L

c.POQ, P = 2

SKUs

123Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall.Copyright 2010 Pearson Education, Inc. Publishing as Prentice Hall..

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